From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Basic Trig
Date: 30 Dec 1999 17:32:35 GMT
Newsgroups: sci.math
Keywords: why doesn't a Taylor series converge?

In article <s6lgjh4o5k233@corp.supernews.com>,
Mike Lugo <mlugo@thelabelguy.com> wrote:
>
>You don't have arcsin?  Well, here's an approximation:
>1*x + 1/6*x^3 + 3/40*x^5 + 5/112*x^7 + 35/1152*x^9 + 63/2816*x^11 +
>231/13312*x^13
>(a Taylor series)
>which has six-place accuracy on |x|<.5, five-place on |x|<.6, four-place on
>|x|<.7, three-place on |x|<.8, and two-place on |x|<.9.  Or you could just
>use a trig table, which I'm sure can be found somewhere (though I'm not sure
>where).

Of course, there's no reason to think the Taylor polynomial is the best
polynomial of a given degree. Welcome to Numerical Analysis.

>Can anyone give me the general term of the series above,

Sure. (d/dx) arcsin(x) = (1-x^2)^(-1/2) = Sum( (-1)^n binomial(-1/2,n) x^(2n) )
Then integrate term by term. You may prefer to replace the binomial
coefficients with suitable expressions involving integers; a little play give
    arcsin(x) = Sum( binomial(2n,n)/(2n+1)/2^(2n) * x^(2n+1) )

>and/or enlighten me on why it seems to take so many terms to get a good 
>approximation near |x|=1?

Look up the error estimates for Taylor series. Basically the problem is
that the derivative of arcsin(x) has a pole at x=1, so that _its_ Taylor
series has to converge slowly. Put another way, it's somehow the presence
of the pole which causes the terms of the Taylor series to decrease
slowly (on the order of  n^(3/2) here) which of course means slow convergence
of the whole series.

dave